To find all non-empty subsets of a collection element

An existing set of n elements to find all the subsets of the set S?

For example: The collection s contains three elements {A, B, C}, then all its subsets are: {} (empty set), {a}, {B}, {C}, {A, b}, {A, c}, {B, C} and {A, B, C}.

it first use the idea of bit operation to solve, concrete method: Use the 2 binary bit to mark whether an element in the collection is selected, 1 is selected, and 0 is not selected. For example, all subsets of the collection {A, B, c} can be represented as follows:

{A} 0 0 1

{b} 0 1 0

{C} 1 0 0

{A, B} 0 1 1

{A, C} 1 0 1

{b, c} 1 1 0

{A, B, c} 1 1 1

From the above analysis can also be seen that a set containing N elements of 2^n-1 a non-empty set, it is very easy to think of the method is to traverse all the integers 1~2^n-1, and into the binary, according to the above thinking output all subsets.

The specific implementation is as follows:

public class Subset {/** * @param args */public static void main (string[] args) {//TODO auto-generated method stublist&lt ; list<string>> result = new arraylist<list<string>> ();//Store all subsets list<string> List = new Arraylist<string> (), List.add ("1"), List.add ("2"), List.add ("3"), List.add ("5");//Gets all the subsets of the elements in the list collection, and deposited in the result set Getsubset (List,result); for (list<string> Li:result) {for (String String:li) {System.out.print ( string+ "");} System.out.println ();}} private static void Getsubset (list<string> sourceset, list<list<string>> result) {//TODO Auto-generated method Stubint n = sourceset.size ();//num element has 2^num-1 non-empty set int num = (int) Math.pow (2, N); for (int i = 1; I & Lt Num i++) {String binary = integer.tobinarystring (i); int size = Binary.length ();//system.out.println ("size" +size); for (int k = 0; K < n-size; k++) {binary = "0" +binary;} SYSTEM.OUT.PRINTLN ("binary" +binary); list<string> set = new Arraylist<string> ();//system.ouT.println (Binary.length ()); for (int index = 0; index < sourceset.size (); index++) {if (Binary.charat (index) = = ' 1 ') {Set . Add (Sourceset.get (index));}} /*for (String string:set) {System.out.print (string+ "");} System.out.println (); */result.add (set);}}}

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To find all non-empty subsets of a collection element